18 Thermal physics Answers - Hodder Education

18 Thermal physics Answers

© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019

Page 329 Test yourself on prior knowledge

1

2 Conduction involves the transfer of the vibration of particles from hot nearest neighbour to

colder nearest neighbours, and convection involves the translation of hotter, faster-moving

particles from somewhere hot to somewhere colder.

3 Particles of water moving fast, and perpendicular to the surface of the tea, can escape the

surface, lowering the overall average kinetic energy of the water particles in the tea, lowering its

temperature.

Page 331 Test yourself

1 Thermodynamics deals with the macroscopic (large-scale) behaviour of a system, in terms of

energies involved with the system, such as the internal energy, the thermal energy and the work

done on or by the system.

2 Increasing its temperature (heating it) OR by doing work on it (compressing it).

3 U = Q – W

so W = Q – U

= 1247 kJ – 1864 kJ

= –617 kJ

The negative sign indicates that, rather than the gas doing work, work is done on the gas.



Page 334–335 Activity

Measuring the specific heat capacity of a metal block

1

2 From graph, T = 44.0 – 16.4 = 27.6°C

3 Specific heat capacity is calculated from Q = mc, where Q is the amount of thermal energy

supplied.

If we assume that the experiment is 100 % efficient, and all the electrical energy supplied from

the heater is converted into thermal energy of the copper block then:

Q = VIt = 12.0 V × 4.0 A × 180 s = 8640 J

The mass of the block, m = 0.814 kg and = 27.6 K so:

c = 𝑄

𝑚 × ∆𝜃

= 8640 J

0.814 kg × 27.6 K

= 384.6 J kg-1 K-1 ≈ 380 J kg-1 K-1 (2 sf)

4 Efficiency of transfer of electrical energy into thermal energy will never be 100 %. Some thermal

energy will be lost to the surroundings.

5 Repeating the experiment would give a range of different values for c. These could be averaged

and the range used to estimate an uncertainty in the value.



6

Error Random or systematic Way that the error could be removed

Measuring the time using a stopwatch

Systematic Use datalogger with appropriate sensor to measure both temperature and time.

Measuring the pd using a voltmeter

Systematic More precise voltmeter. Repeated readings and averaging.

Measuring the current using an ammeter

Systematic More precise ammeter. Repeated readings and averaging.

Measuring the mass of the block using an electronic balance

Systematic More precise balance. Repeated readings and averaging.

Measuring the temperature of the block using a thermometer

Systematic More precise thermometer. Repeated readings and averaging.

The thermal energy escaping to the surroundings

Random Full thermal lagging of the block.

The residual heat left in the heater after it has been switched off

Systematic Use of calorimetric cooling curve technique. Repeated readings and averaging.


4 a) Q = P × t

= 750 W × 150 s = 112 500 J ≈ 110 000 J (2 sf)

= Q/(mc)

= 112 500 / (0.42 kg × 4200 J kg-1 K-1)

= 63.8 K

Final temperature of tea = 17°C + 63.8°C = 80.8°C ≈ 81°C (2 sf)

(Note: 1 degree Celsius is equal to 1 Kelvin.)

b) Some of Q is used changing the state of the water from liquid to gas, so not all the thermal

energy goes into heating all the water in the tea. This will reduce the final temperature of the

tea.

c) = 80.8°C – 71.0°C = 9.8°C

So Q = mc

= 0.42 kg × 4200 J kg-1 K-1 × 9.8°C

= 17 287.2 J = 17 000 J (to 2 sf)

d) m = Q / (c)

= 17 287.2 J / (4000 J kg-1 K-1 × (71.0 – 5.5))

= 0.066 kg



5 a) m = × V

= 1.2 kg m-3 × 24 000 m3

= 28 800 kg

b) = 16°C – 5.0°C = 11°C

So Q = mc

= 28 800 kg × 1000 J kg-1 K-1 × 11°C

= 3.17 × 108 J = 3.2 × 108 J (to 2 sf)

c) Total power = 14.7 kW × 4 = 58 800 W

So t = Q/P

= 3.17 × 108 J / 58 800 W

= 5391 s = 90 mins (to 2 sf)

d) Some thermal energy will be lost via conduction through the material of the dome.

6 a) m = × V

= 1 g cm-3 × (240 × 60 × 60) cm3

= 864 kg ≈ 860 kg (2 sf)

b) Q = mc

= 864 kg × 4200 J kg-1 K-1 × (25.5 – 9.5)°C

= 58 060 800 J

c) Rate of fall of temperature = rate heater causes temperature to rise when functioning

∆𝑡=

𝑄

𝑚𝑐∆𝑡=

𝑃

𝑚𝑐

= 100 W / (864 kg × 4200 J kg-1 K-1)

= 2.75 × 10-5 K s-1

2.75 × 10-5 K s-1 × 3600 s = 0.099 K hr-1 0.1°C hr-1.

d) k = ∆𝜃

∆𝑡⁄

𝜃𝑤−𝜃𝑠 =

0.099 K hr−1

(25.5−15.0)°C = – 0.0094 hr-1

e) ∆𝜃

∆𝑡 = – 0.0094 hr-1 × (25.5 – 8)°C = 0.1645°C hr-1

Pages 338–339 Test yourself

7 Q = mlf

= 12.5 × 10-3 kg × 334 × 10-3 J kg-1

= 4175 J ≈ 4180 J (3 sf)



8 Q = mlf

= 4.2 × 10-3 kg × 23 × 103 J kg-1

= 96.6 J

So time taken, t = Q/P = 96.6 J / 18 W = 5.36 s ≈ 5.4 s (2 sf)

9 a) Q = mc

= 1.5 kg × 4200 J kg-1 K-1 × (100 – 20)°C

= 5.04 × 105 J

b) t = Q/P

= 5.04 × 105 J / 2.7 × 103 W

= 186.6 s (190 s to 2 sf)

c) m = Q/lv = (P × t) / lv

= (2.7 × 103 W × 25 s) / 2260 × 103 J kg-1

= 0.030 kg

10 a) V1I1t = m1lv + E and V2I2t = m2lv + E

so V1I1t − m1lv = V2I2t − m2lv

m2lv − m1lv = V2I2t − V1I1t

lv = (V2I2− V1I1)t

m2− m1

b) lv = (12.00 V × 3.00 A−8.00 V × 2.41 A) ×600 s

(10.3 × 10−3−5.8 × 10−3) kg

= 2.23 × 106 J kg-1 = 2.2 × 106 J kg-1 (2 sf)

11 a) A suitable diagram showing a test tube containing stearic acid, inside a small water bath with

thermometer probes in the stearic acid and the water bath.

b) Thermal energy lost by the water = mc

= 25 × 10-3 kg × 4200 J kg-1K-1 × (95 – 18) oC = 8085 J in 600 s

So P, rate at which energy is lost is 8085 J / 600 s = 13 W

c) i) c = 𝑃𝑡

𝑚∆𝜃

Using readings from portion of graph over which stearic acid is solid:

c = 13 W ×(600−300) s

4 × 10−3g ×(65−20)℃ = 21667 J kg-1 K-1

ii) lf = 𝑃𝑡

𝑚

Using the portion of the graph over which the temperature of the stearic acid is almost

constant:

lf = 13 W ×(245−180) s

4 × 10−3g = 211250 J kg-1



Pages 339–340 Required practical 8 Investigating Boyle’s (constant temperature) law

1

pressure, p, (× 105 Pa); (±0.01 × 105 Pa)

volume, V, (cm3); (±0.5 cm3)

1/V, (cm-3) p x V,

(× 105 Pa cm)

1.0 30.5 0.033 30.5

1.5 20.0 0.050 30.0

2.0 15.5 0.065 31.0

2.5 12.0 0.083 30.0

3.0 10.0 0.100 30.0

3.5 9.0 0.111 31.5

2

3

0

5

10

15

20

25

30

35

0 0.5 1 1.5 2 2.5 3 3.5 4

Vo

lum

e V

(cm

3)

Pressure p (x105 Pa)

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 0.5 1 1.5 2 2.5 3 3.5 4

Vo

lum

e V

(cm

3)

Pressure p (x105 Pa)



4 Excel calculates the gradient of the graph of 1/V against p to be 0.0324 cm-3/(× 105 Pa)

This is 1/k, so k = 30.7 × 105 Pa cm3 = 3.09 N m

The largest uncertainty on the graph gives a value of about 6%, therefore the uncertainty in k is

0.19 N m.

5 The value of V is also dependent on the temperature of the apparatus. Compressing the air does

work on the gas increasing its internal energy and hence the temperature. Allowing the

apparatus to come to thermal equilibrium reduces this systematic error.

Page 342 Required practical 8

Investigation of Charles’ law for a gas

1

2 Examining the graph gives a value of absolute zero in the range of –265 to – 275°C.


12 p1 = 2.5 MPa; V1 = 6.0 × 10-3 m3; T1 = 273.15 + 5 oC = 278.15 K

p2 = 0.1 MPa; V2 = ?; T2 = 273.15 + 20 oC = 293.15 K

𝑝1𝑉1

𝑇1 =

𝑝2𝑉2

𝑇2

V2 = 𝑝1𝑉1𝑇2

𝑇1𝑝2

= 2.5 MPa × 6.0 × 10−3m3 ×293.15 K

278.15 K × 0.1 MPa

= 0.158 m3

13 If p remains constant, VT, so if V doubles, T doubles. T is in Kelvin.

24°C = 297.15 K, hence T2 must be 594.3 K = 321.15°C.



14 a) The volume remains constant, so pT and

𝑝1

𝑇1 =

𝑝2

𝑇2

T2 = 𝑝2𝑇1

𝑝1 =

2.4 × 105Pa ×(16+273.15)°C

1.5 × 105Pa

= 462.54 K ≈ 190°C (2 sf)

b) The water will boil, producing a large quantity of steam, increasing the pressure at a greater

rate.

15 The temperature remains constant, so the air obeys Boyle’s Law where:

P1V1 = p2V2

V2 = 𝑝1𝑉1

𝑝2

= (1+2.5) atm × 22 × 10−3 m3

1 atm

= 77 × 10-3 m3

16 p1 = 1.5 x 106 Pa; V1 = 150 cm3; T1 = 273.15 + 100 oC = 373.15 K

p2 = 1.0 x 105 Pa; V2 = ?; T2 = 273.15 + 6.5 oC = 279.65 K

𝑝1𝑉1

𝑇1 =

𝑝2𝑉2

𝑇2

V2 = 𝑝1𝑉1𝑇2

𝑇1𝑝2

= 1.5 × 106Pa × 150 cm3 × 279.65 K

373.15 K × 1.0 × 105 Pa

= 1686 cm3 ≈ 1700 cm3 (2 sf)


17 V = 𝑛𝑅𝑇

𝑝

= 1.5 mol ×8.31 J mol−1K−1 × 312 K

1.7 × 105Pa

= 0.023 m3

18 n = 𝑝𝑉

𝑅𝑇

= 1.2 ×105Pa × 0.85 m3

8.31 J mol−1K−1 ×(18+273) K

= 42.2 moles ≈ 42 moles (2 sf)

19 a) p = 𝑛𝑅𝑇

𝑉

= 0.15 mol × 8.31 J mol−1K−1 ×293 K

8.2 × 10−4m3

= 4.45 × 105 Pa



b) Assuming that the volume of the tyre does not increase:

T = 𝑝𝑉

𝑛𝑅

= 5.45 × 105 Pa × 8.2 ×10−4 m3

0.15 mol ×8.31 J mol−1K−1

= 358.5 K ≈ 85°C (2 sf)

20 a) D

b) A

c) C

21 V = 𝑛𝑅𝑇

𝑝

= 1 mol ×8.31 J mol−1K−1 × 273K

1.01 × 105Pa

= 0.0225 m3

T = 𝑝𝑉

𝑛𝑅

= 1.01 × 105 Pa × 2.45 ×10−2 m3

1 mol ×8.31 J mol−1K−1

= 297.8 K

Quantity Standard temperature and pressure (STP)

Room temperature and pressure (RTP)

Temperature, T/K 273 297.8

Pressure, p/105 Pa

1.01 1.01

Volume, V/m3 2.25 x 10-2 2.45 × 10−2



Pages 346–347 Activity

The ideal gas equation and Mount Kilimanjaro

1 For 1.0 kg of air on the surrounding plains, at a temperature of (273 + 30) K = 303 K and 90 kPa

Volume = mass/density

= 1.0 kg / 1.03 kg m-3

= 0.97 m3

On the surrounding plains:

p1 = 90 kPa; V1 = 0.97 m3; T1 = 303 K

At the summit:

p2 = 50 kPa; V2 = ?; T2 = (273 − 6.8) K = 266 K

Using the ideal gas combined gas law:

𝑝1𝑉1

𝑇1 =

𝑝2𝑉2

𝑇1

V2 = 𝑝1𝑉1𝑇2

𝑇1𝑝2

= 90 kPa × 0.97 m3 × 266 K

303 K ×50 kPa

= 1.53 m3 ≈ 1.5 m3 (2 sf)

This is the volume of 1 kg of air at the top of Mount Kilimanjaro, thus the density of the air at the

summit is:

⍴ = 𝑚

𝑉 =

1.0

1.53 = 0.65 kg m-3

2 On the surrounding plains, using the ideal gas equation, pV = nRT, 1 mole of air has a volume of:

V1 = 𝑛𝑅𝑇1

𝑝1 =

1 × 8.31 J mol−1K−1 × 303 K

90 × 103Pa = 0.028 m3 = 280 litres

At the summit:

V2 = 𝑛𝑅𝑇2

𝑝2 =

1 × 8.31 J mol−1K−1 × 266 K

50 × 103Pa = 0.044 m3 = 440 litres

So, 6 litre capacity lungs contain:

a) (On the plains) no. of oxygen molecules of air in 6 litres

= 0.21 × 6.02 × 1023 × 6 l

280 l = 2.71 × 1021 molecules

b) (At the summit) no. of oxygen molecules of air in 6 litres

= 0.21 × 6.02 × 1023 × 6 l

440 l = 1.72 × 1021 molecules

This is only 63 % of the plains-level value and will lead to some form of altitude sickness.




22 a) F = N∆𝑝

∆𝑡

= 2.6 × 1024 atoms × 1.33 × 10−22kg m s−1

1𝑠

= 345.8 N ≈ 350 N (2 sf)

b) p = 𝐹

𝐴

= 345.8 N

7.4 × 10−4m2

= 467297 Pa ≈ 4.7 × 105 Pa (2 sf)

23 a) Circumference = 2r

r = C/2 = 70/2 = 11.1 cm

Volume, V = 4

3πr3 =

4

3π(11.1 × 10-2 m)3 = 5.73 × 10-3 m3 (3 sf)

Number of moles of gas, n = 𝑝𝑉

𝑅𝑇 =

6.9 × 105Pa ×5.7 × 10−3m3

8.31 J mol−1 × 300 K = 1.58 ≈ 1.6 (2 sf)

Mass of air, m = n × Mm = 1.58 × 29 × 10-3 kg mol-1 = 0.0458 kg

b) Density of air in football, ρ = 𝑚

𝑉 =

0.0458 kg

5.7 × 10−3m3 = 7.99 ≈ 8.0 kg m-3 (2 sf)

p = 1

3(crms)2

crms = √3𝑝

𝜌

= √3 × 6.9 × 105Pa

8.0 kg m−3

= 497 m s-1 ≈ 500 m s-1 (2 sf)

24 a) R = 𝑝𝑉

𝑛𝑇

= 1.01 × 106 Pa × 0.067 m3

3 × 273 K

= 8.26 J mol-1 K-1 ≈ 8.3 J mol-1 K-1 (2 sf)

b) k = 𝑅

𝑁𝐴

= 8.26 J mol−1K−1

6.02 × 1023

= 1.4 × 10−23 J K

c) Ek = 3

2kT

= 3

2 × 1.37 × 10-23 J K-1 × 273 K

= 5.6 × 10-21 J



25 Points needed:

• KE and therefore average velocity of particles increases with T

• Therefore momentum change on each collision each greater

• Molecules take less time to travel between walls/to the wall and back

• Number of collisions per second increases with T

• Momentum change per second depends on momentum change per collision and number of

collisions per second – both of these are greater, increasing the force on the walls, and thus

the pressure

Assumptions:

• All particles are similar

• Newton’s Laws of Motion are obeyed

• Volume of particles << Volume of container

• Motion is random

• All collisions are totally elastic

• No intermolecular forces

26 a) p = 𝑛𝑅𝑇

𝑉 =

𝑀

𝑀𝑚 ×

𝑅𝑇

𝑉

= 42 g

4.0 g ×

8.31 J mol−1K−1 × (273+20)K

3.0 × 10−3m3

= 8.5 × 106 Pa

b) N = nNA = 𝑀

𝑀𝑚 × NA

= 42 g

4.0 g × 6.02 × 1023

= 6.3 × 1024 molecules

c) pV = 1

3Nm(crms)2 crms = √

3𝑝𝑉

𝑁𝑚

= √3 × 8.5 × 106Pa × 3.0 × 10−3m3

6.3 × 1024×4.0 × 10−3

6.02 × 1023kg

= 1352 ms-1 ≈ 1400 ms-1 (2 sf)

d) p decreases as pT

N remains constant

crms decreases as crmsT

Pages 353–356 Practice questions

1 C

2 B

3 A

4 D

5 B



6 A

7 A

8 D

9 C

10 C

11 a) Qi = micii [1]

= 3.0 kg × 440 J kg−1 K−1 × (35 – 31) °C

= 5280 J [1]

b) Qfreeze = mg × lf

= 25 × 10−3 kg × 63 × 103 J kg−1

= 1575 J [1]

c) Qfreeze + Qcooling = Qi

Qcooling = Qi − Qfreeze [1]

= 5280 J – 1575 J = 3705 J [1]

= mgcgg

cg = 3705 J

(25×10−3 kg) × (1064 ℃ − 35 ℃)

= 144 J kg-1 K-1 [1]

d) No thermal energy is lost to the surroundings/specific heat capacity of gold is constant over a

wide temperature range. [1]

12 a) mtct (40 − T) = maca (T − 5) [1]

0.050 kg × 4250 J kg-1 K-1 (40 − T) = 0.12 kg × 900 J kg-1 K-1(T − 5)

8500 − 212.5T = 108T − 540

9040 = 320.5 T [1]

T = 28.2 °C = 28 °C (2 sf) [1]

b) Time to cool mould and tea to 0℃, t = 𝑄

𝑃=

𝑚𝑡𝑐𝑡∆𝜃𝑡 + 𝑚𝑚𝑐𝑚∆𝜃𝑡

𝑃

= (0.05 kg × 4250 J kg−1 K−1 ×28.2 ℃)+(0.12 kg × 900 J kg−1 K−1 × 28.2 ℃)

32 W = 282 s [1]

Time to freeze tea = mtlf

P=

0.05 kg × 3.38×105 J kg−1

32 W = 528 s [1]

Total time to freeze = 282 s + 528 s = 810 s (13.5 minutes) [1]

We have assumed that the mould remains in thermal equilibrium with the tea as it freezes. [1]

13 a) Per second: Δθ =Q

mc=

15×103 J

0.24 kg × 4200 J kg−1 K−1 = 14.8 °C [1]

so output temperature = 5°C + 14.8 °C = 19.8 °C (or 20 °C to 2 sf) [1]



b) t =mc∆θ

P [1]

=0.24 kg × 4200 J kg−1 K−1 × (80−35)℃

15 × 103 W= 3.0 s [1]

14 a) n =pV

RT=

1.4 × 105 Pa × 0.09 m3

8.31 J mol−1 × 285 K= 5.32 moles (or 5.3 to 2 sf) [1]

b) p2 =p1T2

T1 [1]

=1.4 × 105Pa × 363 K

285 K= 1.78 × 105 Pa = 1.8 × 105 Pa (2 sf) [1]

c) Mass of nitrogen in tyre = 0.028 kg mol−1 × 5.32 moles = 0.149 kg [1]

density ρ =𝑚

𝑉=

0.149 kg

0.09 m3 = 1.65 kg m−3 [1]

𝑐𝑟𝑚𝑠 = √3𝑝

𝜌= √

3 × 1.78 × 105 Pa

1.65 kg m−3 = 568.9 m s−1 = 570 m s−1 (2 sf) [1]

d) Similarity: move in random directions/no intermolecular forces/obey Newton’s Laws. [1]

Difference: velocity / kinetic energy / hit walls with different forces. [1]

15 a) [1 mark for correct scale, 1 mark for points plotted correctly, 1 mark for line of best fit through

(0,0)]

b) As pV = nRT, gradient of pT graph = nR/V [1]

So, n = gradient × V/R [1]

= 0.0041 ×0.055 m3

8.31 Jmol-1 =2.7 × 10−5 moles (using Excel calculated gradient) [1]

c) T = 120–128 K as measured from graph intercept at 0.50 atm. [1]

d) Ek = 3

2 kT [1]

= 3

2 × 1.37 × 10−23 J K−1 × 124 K = 2.55 × 10−21 J (or similar, using answer to (c)) [1]



e) U = NEk = nNAEk [1]

= 2.7 × 10−5 × 6.02 × 1023 × 2.55 × 10−21 J = 0.041 J (or similar, using answer to (d)) [1]

16 a) 2 or 3 points from list = [1], 4 or 5 points = [2]

• Molecules have negligible volume

• Collisions are elastic

• Gas cannot be liquefied

• There are no intermolecular forces (apart from during collisions)

• Gas obeys gas laws at all temperatures/pressures [2]

b) n =pV

RT [1]

=2.02 × 105 Pa × 3.3 × 10−4 m3

8.31 J mol−1 × 300 K = 0.027 moles [1]

c) ⍴ = nMm

V [1]

= 0.027 moles × 0.084 kg mol−1

3.3 × 10−4 m3 [1]

= 687 kg m-3 = 6.9 kg m−3 (2 sf) [1]

d) V2 = p1V1T2

T1p2

= 2.02 × 105 Pa × 3.3 × 10−4 m3 × 266 K

300 K × 0.5 × 105 Pa

= 1.18 × 10-3 m3 = 1.2 × 10−3 m3 (2 sf) [1]

Pages 356–357 Stretch and challenge

17 a) pressure × volume = constant, for constant mass and temperature

b) initial pressure in air column = (A + 100) mmHg

where A = atmospheric pressure in mmHg

second pressure in air column = (A − 100) mmHg

The volume of trapped air is proportional to the length of the air column, so applying Boyle’s

Law gives (A + 100) × (400) = (A − 100) × (520)

400A + 40000 = 520A – 52000

A = 92000

120 = 767 mmHg

c) i) force = pressure × area

area = 4R2 = 5.15 × 1014 m2

force = (101 × 103 Pa) × (5.15 × 1014 m2) = 5.2 × 1019 N

ii) W = mg so divide result from part i) by g = 9.8 m s−2

mass = 5.3 × 1018 kg



iii) Number of molecules, N = Avogadro Number × Number of moles

N = 6.02 × 1023 ×5.3 × 1018 kg

0.030 kg mol−1

N = 1.06 × 1044 molecules

iv) Volume, V = mass / density

= 5.3 × 1018 kg

1.2 kg m−3

= 4.4 × 1018 m3

Height, h = volume

surface area of earth

= 4.4 × 1018 m3

4 ×π ×(6.4 × 106 m)2

= 8600 m

d) The air is compressible and the air at the ground is compressed due to the weight of the air

above it. We have used the density of air at ground level for the calculations but it is less at

higher levels so volume is greater than that calculated and hence height is also larger.

e) The value for the mass obtained is good because the weight of the air (and hence the mass) is

independent of whether or not it is compressed, (as long as the height is not so great that g

reduces significantly – reasonable to think this applies here as 200 km < 6400 km).

18 Let T be the original temperature:

Heat gained by thermometer = heat lost by water

20 J K−1 × (50 – 18) °C = (T – 50) °C × 4200 J kg−1 K−1 × 0.25 kg

640 = (T – 50) × 1050

T = 50.0 + (640 / 1050)

= 50.6 °C

19 a) Wind drives water vapour from the clothes. This causes more water to evaporate, which cools

the remaining water in the clothes. (Sufficient cooling will extract sufficient latent heat of

fusion from the water to cause freezing of the water.)

b) If mass M freezes and mass m evaporates:

Heat extracted from M, EM = (333 × 103 J kg−1 × M) J

Energy absorbed by m, Em = (2500 × 103 J kg−1 × m) J

Equating these two expressions:

333M = 2500m

M

m=

2500

333

M

m+M=

2500

2500 + 333= 0.882



20 Let m be the mass of the bullet and v be the speed of the bullet.

80% kinetic energy of bullet = thermal energy heating bullet + thermal energy melting bullet

4

5(1

2mv2) = (m × 0.12 × 103 J kg-1 K-1 × (600 – 320) K) + (m × 21 × 103 K kg-1)

v2 = 5

2(3.36 × 104 + 21 × 103)

v = 370 m s-1

21 a) Thermal energy to raise temperature of water from 20°C to 100°C = m × 4200 J kg−1 K−1 × 80 K

Thermal energy required to boil the water at 100°C = mlv

Thermal energy supplied per sec, P

P = m kg × 4200 J kg−1 K−1 × 80 K

180 s=

m × 4200 × 4

9

In 1200 s, the energy supplied is:

1200 × P = m × lv

P = mlv

1200

Hence mlv

1200=

m × 4200 × 4

9

lv = 2.24 × 106 J kg-1

Assumption: the thermal capacity of the kettle is negligible.

b) i) Atmospheric pressure, p = 1.01 × 105 Pa

ii) Watmosphere = ʃ pdV

= 1.01 × 105 Pa × 0.10 m × π × (0.24 m

2)

2 = 457 J

iii) 100°C

(Because the change has happened slowly allowing thermal equilibrium to be maintained.)

iv) Since the volume of gas has halved while the pressure and temperature remained

constant, half of the gas has condensed to water.

Qc = −U − W = −lvm − Watmosphere

= − (2.24 × 106 J kg−1) × 0.5 × (0.37 × 10−3 kg) – 457 J

= −414 J – 457 J = −871 J

(Negative sign because heat is transferred to the surroundings)

c) i) height reached is given by 1

2mv2 = mgh

but 1

2mv2=

3

2kT

and m = 𝑀𝑚

𝑁𝐴=

0.032 kg

6.02 × 1023

so h = 3kT

2mg =

3

2×

1.38 × 10−23 J K−1 × 283 K

(0.032 kg

6.02 × 1023⁄ ) × 9.81 N kg−1 = 1.1 ×104 m



ii) 3

2kTS =

GmME

RE

TS = 2

3k× (

GmME

RE)

Using ME = 5.97 × 1024 kg and RE = 6.37 × 106 m

= 2 × 6.67 × 10−11 N m2 kg−2 ×(

0.032 kg

6.02 × 1023)× 5.97 × 1024 kg

3 × 1.38 × 10−23 J K−1 × 6.37 × 106 m

= 1.60 × 105 K

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18 Thermal physics Answers - Hodder Education

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